SSC CGL Tier 2 Quant Previous Year Paper (Held On: 15 November 2020)

Given:

$3 ÷ 18 \text{ of } 3 \times 6 + 21 \times 6 ÷ 18 - 3 ÷ 2 + 3 - 3 ÷ 9 \text{ of } 3 \times 9$ = ?

Formula used:

BODMAS rule (Bracket, Of, Division, Multiplication, Addition, Subtraction)

Calculation:

? = 3 ÷ (18 × 3) × 6 + 21 × 6 ÷ 18 - 3 ÷ 2 + 3 - 3 ÷ (9 × 3) × 9

⇒ ? = 3 ÷ 54 × 6 + 126 ÷ 18 - 1.5 + 3 - 3 ÷ 27 × 9

⇒ ? = $\frac{3}{54} \times 6 + 7 - 1.5 + 3 - \frac{3}{27} \times 9$

⇒ ? = $\frac{1}{18} \times 6 + 7 - \frac{3}{2} + 3 - \frac{1}{9} \times 9$

⇒ ? = $\frac{1}{3} + 7 - \frac{3}{2} + 3 - 1$

⇒ ? = $\frac{1}{3} - \frac{3}{2} + 9$

⇒ ? = $\frac{2 - 9 + 54}{6}$

⇒ ? = $\frac{47}{6}$

∴ The correct answer is option (3).

Shortcut Trick

When two medians m₁ and m₂ of a triangle intersect at 90°, the area of the triangle is given by the formula:

Area of ΔABC = (2 ÷ 3) × m₁ × m₂

Area = (2 ÷ 3) × 10.8 × 14.4

Area = 2 × 3.6 × 14.4 = 7.2 × 14.4 = 103.68 cm²

∴ The correct answer is 103.68 cm².

Alternate Method

Given: Median AD = 10.8 cm, Median BE = 14.4 cm, AD ⊥ BE.

Formula Used: Centroid divides medians in 2 : 1 ratio. Area of ΔABC = 3 × Area of ΔGAB.

⇒ G is the centroid of ΔABC.

⇒ AG = (2 ÷ 3) × AD = (2 ÷ 3) × 10.8 = 7.2 cm

⇒ BG = (2 ÷ 3) × BE = (2 ÷ 3) × 14.4 = 9.6 cm

⇒ Since AD ⊥ BE, ∠AGB = 90°.

⇒ Area of ΔGAB = (1 ÷ 2) × AG × BG

⇒ Area of ΔGAB = (1 ÷ 2) × 7.2 × 9.6 = 3.6 × 9.6 = 34.56 cm²

⇒ Total Area of ΔABC = 3 × Area of ΔGAB

⇒ Area of ΔABC = 3 × 34.56 = 103.68 cm²

∴ The correct answer is 103.68 cm².

Additional Information

Centroid Ratio Property

The centroid (G) divides each median into two parts in the ratio 2 : 1 from the vertex to the midpoint of the opposite side.

Area and Medians

The three medians of a triangle divide it into six smaller triangles of equal area.

General Area Formula via Medians

If m₁, m₂, and m₃ are medians, Area of ΔABC = (4 ÷ 3) × √[s_m(s_m−m₁)(s_m−m₂)(s_m−m₃)], where s_m is the semi-sum of medians.

Orthogonal Medians Relation

If medians AD and BE are perpendicular, then the side lengths satisfy the relation AB² + BC² + CA² related properties, specifically 5c² = a² + b² where c is the side BC.

Shortcut Trick

Apply Alligation between Expenditure and Savings percentage changes.

Expenditure change = 20%, Savings change = 50%, Mean (Income) change = 26%.

Ratio of Exp : Sav = (50 − 26) : (26 − 20) = 24 : 6 = 4 : 1.

Total Income = 4 + 1 = 5 units.

Savings percentage (x) = (Savings / Income) × 100 = (1 / 5) × 100 = 20%.

∴ The correct answer is 20%.

Alternate Method

Given:

Income increase = 26%

Expenditure increase = 20%

Savings increase = 50%

Initial Savings = x% of Income

Calculations:

Let original Income = 100

Original Savings = x, Original Expenditure = 100 − x

New Income = 100 + 26% of 100 = 126

New Expenditure = (100 − x) + 20% of (100 − x)

⇒ New Expenditure = 1.20 × (100 − x) = 120 − 1.2x

New Savings = x + 50% of x = 1.5x

Since Income = Expenditure + Savings:

⇒ 126 = (120 − 1.2x) + 1.5x

⇒ 126 = 120 + 0.3x

⇒ 6 = 0.3x

⇒ x = 6 ÷ 0.3

⇒ x = 20

∴ The correct answer is 20%.

Additional Information

Income-Expenditure Formula

Income = Expenditure + Savings. Any percentage change in income is the weighted average of percentage changes in expenditure and savings.

Alligation in Percentage Changes

If two components of a sum change by different percentages, the ratio of the components can be found using alligation: (Sav% change − Inc% change) : (Inc% change − Exp% change).

Base Value Concept

When solving percentage increase/decrease problems, assuming the initial total value (like income) as 100 simplifies the linear equations significantly.

Shortcut Trick

Perform repeated inversion (reciprocal) on 45/53 to extract integer parts.

Step 1: 53 ÷ 45 = 1 (rem 8) → a = 1.

Step 2: 45 ÷ 8 = 5 (rem 5) → b = 5.

Step 3: 8 ÷ 5 = c − 2/5 ⇒ 1.6 = c − 0.4 ⇒ c = 2.

Value = 4(1) − 5 + 3(2) = 4 − 5 + 6 = 5.

∴ The correct answer is 5.

Alternate Method

Given: 45/53 = 1 / (a + 1 / (b + 1 / (c − 2/5)))

⇒ 45/53 = 1 / (53/45)

⇒ 53/45 = 1 + 8/45 (Comparing with a + ..., a = 1)

⇒ 8/45 = 1 / (45/8)

⇒ 45/8 = 5 + 5/8 (Comparing with b + ..., b = 5)

⇒ 5/8 = 1 / (8/5)

⇒ 8/5 = c − 2/5

⇒ 1.6 = c − 0.4

⇒ c = 1.6 + 0.4 = 2

⇒ We have a = 1, b = 5, c = 2

⇒ Value = 4a − b + 3c = 4(1) − 5 + 3(2)

⇒ Value = 4 − 5 + 6 = 5

∴ The correct answer is 5.

Additional Information

Continued Fraction Properties

A rational number can always be expressed as a finite continued fraction using the Euclidean algorithm process.

Reciprocal Rule

To simplify fractions of the form 1/x = y, we can always write x = 1/y to solve for variables in the denominator.

Integer Constraint

When a, b, c are specified as integers, the decomposition must result in whole numbers for these variables at each step of the division.

Shortcut Trick

Ratio 1 = 3 : 4 : 5 (Sum = 12)

Ratio 2 = 1/3 : 1/4 : 1/5 → 20 : 15 : 12 (Sum = 47)

To compare, equate sums: LCM(12, 47) = 564 units.

Lowest share son (Ratio 1) = 3/12 × 564 = 141 units.

His share (Ratio 2) = 20/47 × 564 = 240 units.

Difference = 240 − 141 = 99 units → ₹1188 ∴ 1 unit = ₹12. Total = 564 × 12 = ₹6768.

∴ The correct answer is ₹6768.

Alternate Method

Given:

Ratio 1 = 3 : 4 : 5

Ratio 2 = 1/3 : 1/4 : 1/5

Difference in share of the first son = ₹1188

Formula:

Individual Share = (Individual Ratio / Sum of Ratios) × Total Amount

Calculations:

⇒ Let the total amount be T.

⇒ Sum of Ratio 1 = 3 + 4 + 5 = 12

⇒ Initial share of the son with lowest share (3) = (3/12) × T = T/4

⇒ Ratio 2 = 1/3 : 1/4 : 1/5 = 20 : 15 : 12 (by multiplying by LCM 60)

⇒ Sum of Ratio 2 = 20 + 15 + 12 = 47

⇒ New share of the same son = (20/47) × T

⇒ According to the question: (20T / 47) − (T / 4) = 1188

⇒ (80T − 47T) / 188 = 1188

⇒ 33T / 188 = 1188

⇒ T = (1188 × 188) / 33

⇒ T = 36 × 188 = 6768

∴ The correct answer is ₹6768.

Additional Information

Ratio Conversion

To convert a fractional ratio a/b : c/d to integers, multiply all terms by the LCM of denominators (b, d).

Inverse Ratio

The inverse ratio of a : b : c is 1/a : 1/b : 1/c, which is used in problems involving speed/time or efficiency/days.

Proportional Division

If an amount A is divided in ratio a : b : c, the shares are [a/(a+b+c)]×A, [b/(a+b+c)]×A, and [c/(a+b+c)]×A.

Shortcut Trick

Substitute θ = 45° in the given expression.

sec 45° = √2, tan 45° = 1, cot 45° = 1, sin 45° = 1/√2, cosec 45° = √2.

Numerator = (√2)² (2 + 1² + 1²) ÷ ((1/√2)² − 1²) = 2(4) ÷ (1/2 − 1) = 8 ÷ (−1/2) = −16.

Denominator = ((√2)² + (√2)²) (1 + 1²)² = (2 + 2) (2)² = 4 × 4 = 16.

Final Value = Numerator ÷ Denominator = −16 ÷ 16 = −1.

∴ The correct answer is − 1.

Alternate Method

Given: Expression = [sec²θ (2 + tan²θ + cot²θ) ÷ (sin²θ − tan²θ)] / [(cosec²θ + sec²θ)(1 + cot²θ)²]

Formula Used: 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ, and sin²θ + cos²θ = 1.

⇒ Numerator Component 1: 2 + tan²θ + cot²θ = (1 + tan²θ) + (1 + cot²θ) = sec²θ + cosec²θ.

⇒ Numerator Component 2: sin²θ − tan²θ = sin²θ − sin²θ/cos²θ = sin²θ(1 − 1/cos²θ) = sin²θ((cos²θ − 1)/cos²θ) = −sin⁴θ / cos²θ.

⇒ Total Numerator = [sec²θ (sec²θ + cosec²θ)] ÷ [−sin⁴θ / cos²θ].

⇒ Denominator = (cosec²θ + sec²θ) × (cosec²θ)² = (sec²θ + cosec²θ) × cosec⁴θ.

⇒ Final Expression = [sec²θ (sec²θ + cosec²θ) / (−sin⁴θ / cos²θ)] / [(sec²θ + cosec²θ) cosec⁴θ].

⇒ Canceling common terms: Final Value = [sec²θ / (−sin⁴θ / cos²θ)] / cosec⁴θ.

⇒ Final Value = [(1/cos²θ) × (−cos²θ / sin⁴θ)] / (1/sin⁴θ) = [−1 / sin⁴θ] / [1 / sin⁴θ] = −1.

∴ The correct answer is − 1.

Additional Information

Fundamental Identities

The three basic identities are sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Secant and Cosecant Sum

sec²θ + cosec²θ = 1/cos²θ + 1/sin²θ = (sin²θ + cos²θ) / (sin²θ cos²θ) = 1 / (sin²θ cos²θ) = sec²θ × cosec²θ.

Reciprocal Relations

sin θ = 1/cosec θ, cos θ = 1/sec θ, and tan θ = 1/cot θ.

Given:

Age of person between 10 to 40 years = 14 + 20 + 24

Age of person between 50 to 80 years = 32 + 26 + 22

Formula used:

Required Percentage = (Age b/w 50 - 80 – Age b/w 10 - 40)/Age b/w 50 - 80 × 100

Calculation:

Age of person between 10 to 40 years = 14 + 20 + 24

⇒ Age b/w 10 - 40 = 58

Age of person between 50 to 80 years = 32 + 26 + 22

⇒ Age b/w 50 - 80 = 80

Required Percentage = (Age b/w 50 - 80 – Age b/w 10 - 40)/Age b/w 50 - 80 × 100

⇒ Required percentage = (80 – 58)/80 × 100

⇒ Required percentage = 22/80 × 100

⇒ Required percentage = 27.5%

∴ The required percentage is 27.5%.

Given:

(x + 20)% of 250 = 125% of (x% of 220)

Formula used:

a% of b = $\frac{a \times b}{100}$

Percentage less = $\frac{Base - Value}{Base} \times 100$

Calculation:

$\frac{x + 20}{100} \times 250 = \frac{125}{100} \times \frac{x \times 220}{100}$

⇒ $2.5 \times (x + 20) = 1.25 \times 2.2x$

⇒ $2.5x + 50 = 2.75x$

⇒ $0.25x = 50$

⇒ $x = 200$

10% of (x + 50) = 10% of (200 + 50) = 25

15% of x = 15% of 200 = 30

⇒ Required % = $\frac{30 - 25}{30} \times 100$

⇒ $\frac{5}{30} \times 100 = \frac{100}{6} = 16 \frac{2}{3}\%$

∴ The correct answer is option (1).

Shortcut Trick

Difference in profit percentage = 30% − 20% = 10%

This 10% difference corresponds to the price increase of ₹8.

Cost Price (100%) of 1 article = (8 ÷ 10) × 100 = ₹80

Cost Price of 16 articles = 16 × 80 = ₹1280

∴ The correct answer is ₹1280.

Alternate Method

Given:

Initial profit = 20%

Final profit = 30%

Increase in Selling Price (SP) = ₹8

Number of articles = 16

Formula Used:

SP = CP × (100 + Profit%) ÷ 100

Calculations:

Let the Cost Price (CP) of one article be x.

⇒ Initial SP₁ = 1.20x

⇒ New SP₂ = 1.30x

⇒ According to question: SP₂ − SP₁ = 8

⇒ 1.30x − 1.20x = 8

⇒ 0.10x = 8

⇒ x = 8 ÷ 0.10 = 80

⇒ CP of 1 article = ₹80

⇒ CP of 16 articles = 16 × 80

⇒ Total CP = ₹1280

∴ The correct answer is ₹1280.

Additional Information

Profit and Loss Percentage

Profit % = (Profit ÷ CP) × 100. Profit and loss are always calculated on the Cost Price unless specified otherwise.

Successive Percentage Change

When there are two successive changes of a% and b%, the effective change is [a + b + (a × b) ÷ 100] %.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). SP = MP − Discount. The relation is CP × (100 + P%) = MP × (100 − D%).

Shortcut Trick

In any triangle, the angle formed at the in-center by two vertices is 90° + half of the third vertex angle.

Formula: ∠QOP = 90° + (∠R ÷ 2)

Given ∠R = 42°

∠QOP = 90° + (42° ÷ 2) = 90° + 21° = 111°

∴ The correct answer is 111°.

Alternate Method

Given: O is the in-center of ΔPQR and ∠R = 42°.

Formula: Angle at in-center = 90° + (Opposite Angle ÷ 2)

In ΔPQR, the sum of internal angles is 180°.

⇒ ∠P + ∠Q + ∠R = 180°

⇒ ∠P + ∠Q = 180° − 42° = 138°

The in-center (O) is the intersection of angle bisectors.

⇒ In ΔQOP: ∠QOP + ∠OPQ + ∠OQP = 180°

⇒ ∠QOP + (∠P ÷ 2) + (∠Q ÷ 2) = 180°

⇒ ∠QOP = 180° − [(∠P + ∠Q) ÷ 2]

⇒ ∠QOP = 180° − (138° ÷ 2) = 180° − 69°

⇒ ∠QOP = 111°

∴ The correct answer is 111°.

Additional Information

In-center Property

The in-center is the point of intersection of the internal angle bisectors of a triangle and is equidistant from all sides.

Circumcenter Angle

The angle subtended by a side at the circumcenter is twice the angle at the opposite vertex: ∠QCP = 2 × ∠R.

Orthocenter Angle

The angle subtended at the orthocenter (H) and the opposite vertex angle are supplementary: ∠QHP = 180° − ∠R.

Ex-center Angle

The angle at the ex-center (E) opposite to vertex R is given by: ∠QEP = 90° − (∠R ÷ 2).

Shortcut Trick

Total sum of ages 4 years hence = 52 + (4 × 2) = 60 years.

Ratio 4 years hence (Son : Father) = 1 : 4.

Total parts = 1 + 4 = 5 parts ⇒ 5 parts = 60 ⇒ 1 part = 12.

Ages 4 years hence: Son = 12 years, Father = 48 years.

Ages 10 years from now (6 years after the 4-year mark): Son = 12 + 6 = 18, Father = 48 + 6 = 54.

Required Ratio = 18 : 54 = 1 : 3.

∴ The correct answer is 1 : 3.

Alternate Method

Given:

Sum of present ages (F + S) = 52 years

Let Son's present age = x years, Father's present age = (52 − x) years

Calculations:

According to the question, 4 years hence:

⇒ (x + 4) = (1/4) × ((52 − x) + 4)

⇒ 4 × (x + 4) = 56 − x

⇒ 4x + 16 = 56 − x

⇒ 4x + x = 56 − 16

⇒ 5x = 40

⇒ x = 8

Present ages: Son = 8 years, Father = 52 − 8 = 44 years.

Ages 10 years from now:

⇒ Son's age = 8 + 10 = 18 years

⇒ Father's age = 44 + 10 = 54 years

⇒ Required Ratio = 18 : 54 = 1 : 3

∴ The correct answer is 1 : 3.

Additional Information

Sum of Ages Change

If the sum of ages of 'n' people is S, then after 't' years, the new sum will be S + (n × t).

Constant Age Difference

The difference between the ages of two individuals remains constant regardless of the number of years that pass.

Ratio and Scaling

If a ratio is given as a : b and the sum is S, then the values are (a / (a+b)) × S and (b / (a+b)) × S.

Shortcut Trick

Assume θ = 45°, so sinθ = 1/√2, cosθ = 1/√2, tanθ = 1, cotθ = 1, secθ = √2, cosecθ = √2.

Substitute in expression: (1/(1/√2) − 1/(1/√2)) + 1/(√2 − 1) − 1/(√2 + 1)

⇒ (√2 − √2) + (√2 + 1) − (√2 − 1) = 0 + 2 = 2.

Check options at 45°: Option (4) sec45° × cosec45° = √2 × √2 = 2.

This matches the value of the expression.

∴ The correct answer is secθ cosecθ.

Alternate Method

Given: (1/cosθ − 1/sinθ) + 1/(cosecθ − cotθ) − 1/(secθ + tanθ)

Formulas: cosec²θ − cot²θ = 1 and sec²θ − tan²θ = 1

⇒ Expression = (secθ − cosecθ) + (cosecθ + cotθ) − (secθ − tanθ)

⇒ Expression = secθ − cosecθ + cosecθ + cotθ − secθ + tanθ

⇒ Expression = cotθ + tanθ

⇒ Expression = (cosθ/sinθ) + (sinθ/cosθ)

⇒ Expression = (cos²θ + sin²θ) / (sinθ cosθ)

⇒ Expression = 1 / (sinθ cosθ) = secθ cosecθ

∴ The correct answer is secθ cosecθ.

Additional Information

Trigonometric Conjugates

1/(secθ + tanθ) = secθ − tanθ and 1/(cosecθ − cotθ) = cosecθ + cotθ because their products result in 1.

Pythagorean Identities

sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.

Reciprocal Relations

secθ = 1/cosθ, cosecθ = 1/sinθ, tanθ = sinθ/cosθ, and cotθ = cosθ/sinθ.

Shortcut Trick

For 40 km, Time difference (A − B) = 2 hours.

For 80 km at 2 × Speed A, A's time is the same as for 40 km at Speed A.

New difference for 80 km: Time A (for 40 km) − Time B (for 80 km) = 1.5 hours.

Difference in B's times for 80 km and 40 km = 2 − 1.5 = 0.5 hours.

B covers the extra 40 km in 0.5 hours ⇒ Speed of B = 40 ÷ 0.5 = 80 kmph.

Time for 90 km = 90 ÷ 80 = 1 1/8 hours.

∴ The correct answer is 1 1/8 hours.

Alternate Method

Given:

Distance 1 = 40 km, Distance 2 = 80 km, Time difference 1 = 2 hours, Time difference 2 = 1.5 hours.

Calculations:

Let S_A and S_B be the speeds of A and B respectively.

⇒ Case 1: 40/S_A − 40/S_B = 2 ... (i)

⇒ Case 2: 80/(2S_A) − 80/S_B = 1.5

⇒ 40/S_A − 80/S_B = 1.5 ... (ii)

⇒ From (i), 40/S_A = 40/S_B + 2

⇒ Substitute this in (ii): (40/S_B + 2) − 80/S_B = 1.5

⇒ 2 − 1.5 = 80/S_B − 40/S_B

⇒ 0.5 = 40/S_B

⇒ S_B = 40 ÷ 0.5 = 80 kmph

⇒ Time taken by B to cover 90 km = 90 ÷ 80

⇒ Time = 9/8 = 1 1/8 hours.

∴ The correct answer is 1 1/8 hours.

Additional Information

Basic Speed-Distance Formula

Speed = Distance ÷ Time | Distance = Speed × Time | Time = Distance ÷ Speed

Relative Speed

When two objects move in the same direction, relative speed = S₁ − S₂. When they move in opposite directions, relative speed = S₁ + S₂.

Average Speed

For the same distance covered at speeds x and y, Average Speed = (2 × x × y) ÷ (x + y).

Inverse Proportion

For a constant distance, speed is inversely proportional to time (S ∝ 1/T). If speed doubles, time is halved.

Shortcut Trick

Ratio of Gain : Loss = 120 : 100 = 6 : 5.

Difference between SP₁ and SP₂ = 123.40 − 108 = 15.40.

Total ratio units (6 + 5) = 11 units ⇒ 1 unit = 15.40 ÷ 11 = 1.40.

Loss = 5 units = 5 × 1.40 = 7 ⇒ CP = 108 + 7 = 115.

New Gain = 120.75 − 115 = 5.75 ⇒ Gain % = (5.75 ÷ 115) × 100 = 5%.

∴ The correct answer is Gain 5%.

Alternate Method

Given:

Selling Price 1 (SP₁) = ₹ 123.40 (Gain case)

Selling Price 2 (SP₂) = ₹ 108 (Loss case)

New Selling Price (SP₃) = ₹ 120.75

Formula Used:

Profit = SP − CP | Loss = CP − SP | Gain% = (Gain ÷ CP) × 100

⇒ Let the loss incurred at ₹ 108 be x.

⇒ Profit at ₹ 123.40 = x + 20% of x = 1.2x.

⇒ CP = SP₂ + Loss = 108 + x.

⇒ CP = SP₁ − Gain = 123.40 − 1.2x.

⇒ 108 + x = 123.40 − 1.2x

⇒ x + 1.2x = 123.40 − 108

⇒ 2.2x = 15.40

⇒ x = 15.40 ÷ 2.2 = 7.

⇒ CP = 108 + 7 = ₹ 115.

⇒ New Profit = SP₃ − CP = 120.75 − 115 = ₹ 5.75.

⇒ Gain % = (5.75 ÷ 115) × 100 = (575 ÷ 115)% = 5%.

∴ The correct answer is Gain 5%.

Additional Information

Profit and Loss Percentage

Profit % = (Profit/CP) × 100 and Loss % = (Loss/CP) × 100. Always calculate percentage on Cost Price unless stated otherwise.

Successive Percentage Change

If an article is sold through multiple hands at a, b, and c percents, effective multiplier = (1 ± a/100) × (1 ± b/100) × (1 ± c/100).

Relation between CP and SP

SP = CP × [(100 ± P or L%)/100]. Use this to jump directly between CP and SP.

Shortcut Trick

The angle between the altitude and the internal angle bisector from the same vertex is half the absolute difference of the other two angles.

Formula: ∠SPT = 1/2 |∠Q − ∠R|

Given ∠Q = 84° and ∠R = 48°.

∠SPT = 1/2 |84° − 48°|

∠SPT = 1/2 × 36° = 18°

∴ The correct answer is 18°.

Alternate Method

Given: In ΔPQR, ∠Q = 84°, ∠R = 48°, PS ⊥ QR, and PT is the bisector of ∠P.

Formula Used: Sum of angles in a triangle = 180°; Angle bisector divides the angle into two equal parts.

⇒ In ΔPQR, ∠P + ∠Q + ∠R = 180°

⇒ ∠P + 84° + 48° = 180°

⇒ ∠P = 180° − 132° = 48°

⇒ Since PT is the angle bisector of ∠P, ∠QPT = ∠P ÷ 2 = 48° ÷ 2 = 24°

⇒ In right-angled ΔPQS, ∠PSQ = 90° and ∠PQS = 84°

⇒ ∠QPS = 180° − (90° + 84°) = 180° − 174° = 6°

⇒ ∠SPT = ∠QPT − ∠QPS

⇒ ∠SPT = 24° − 6° = 18°

∴ The correct answer is 18°.

Additional Information

Altitude-Bisector Relation

In any triangle, the angle between the altitude and the internal angle bisector drawn from the same vertex is equal to half the positive difference of the other two angles of the triangle.

Sum of Angles Property

The sum of all internal angles of any triangle is always 180°. This is used to find the third angle when two angles are known.

Angle Bisector Theorem

An angle bisector divides the opposite side into two segments that are proportional to the other two sides of the triangle: QT/TR = PQ/PR.

Shortcut Trick

In a right-angled ΔABC (at A), the length of the perpendicular DE from the angle bisector AD to a side is given by the side of the square rule.

Length of DE = (AB × AC) ÷ (AB + AC)

DE = (10 × 15) ÷ (10 + 15)

DE = 150 ÷ 25 = 6 cm.

∴ The correct answer is 6 cm.

Alternate Method

Given: AB = 10 cm, AC = 15 cm, ∠A = 90°, AD is the bisector of ∠A, and DE ⊥ AC.

Formula: In ΔABC, if AD is the angle bisector of ∠A, then ∠BAD = ∠DAC = ∠A/2.

⇒ Since AD is the bisector of ∠A (90°), ∠DAC = 45°.

⇒ In right-angled ΔADE, ∠AED = 90° and ∠DAE = 45°.

⇒ ∠ADE = 180° − (90° + 45°) = 45°.

⇒ Since ∠DAE = ∠ADE = 45°, ΔADE is an isosceles right triangle, so AE = DE.

⇒ Let DE = AE = x cm. Then CE = AC − AE = 15 − x cm.

⇒ In ΔCDE and ΔCAB, ∠C is common and ∠CED = ∠CAB = 90°.

⇒ ΔCDE ≈ ΔCAB (By AA similarity criterion).

⇒ DE ÷ AB = CE ÷ AC

⇒ x ÷ 10 = (15 − x) ÷ 15

⇒ 15x = 10(15 − x)

⇒ 15x = 150 − 10x

⇒ 25x = 150

⇒ x = 150 ÷ 25 = 6 cm.

∴ The correct answer is 6 cm.

Additional Information

Internal Angle Bisector Theorem

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. BD ÷ DC = AB ÷ AC.

Length of Angle Bisector

In any triangle ABC, the length of the internal angle bisector AD is given by (2bc cos(A/2)) ÷ (b + c).

Inradius of a Right Triangle

The radius of the incircle of a right-angled triangle is r = (Perpendicular + Base − Hypotenuse) ÷ 2.

Shortcut Trick

Side of square base (s) = 10 cm, Height (h) = 10 cm.

Slant height (l) = √(h² + (s/2)²) = √(10² + 5²) = √125 = 5√5 cm.

Lateral Surface Area = 2 × s × l = 2 × 10 × 5√5 = 100√5 cm².

∴ The correct answer is 100√5 cm².

Alternate Method

Given: Side of square base (s) = 10 cm, Height of pyramid (h) = 10 cm.

Formula Used: Lateral Surface Area = 1/2 × Perimeter of base × Slant height (l)

⇒ Distance from center of base to midpoint of side = 10 ÷ 2 = 5 cm.

⇒ Slant height (l) = √(Height² + 5²) = √(10² + 5²)

⇒ l = √(100 + 25) = √125 = 5√5 cm.

⇒ Perimeter of square base = 4 × 10 = 40 cm.

⇒ Lateral Surface Area = 1/2 × 40 × 5√5

⇒ Lateral Surface Area = 20 × 5√5 = 100√5 cm².

∴ The correct answer is 100√5 cm².

Additional Information

Total Surface Area (TSA)

The total surface area of a pyramid is the sum of the lateral surface area and the area of the base. TSA = LSA + Base Area.

Volume of a Pyramid

The volume is one-third of the product of the base area and the vertical height. Volume = 1/3 × Base Area × Height.

Slant Edge Formula

The slant edge (e) can be calculated using the slant height (l) and half the side (s/2): e = √(l² + (s/2)²).

Shortcut Trick

Rate for 8 months = (15% ÷ 12) × 8 = 10%.

Number of periods in 2 years (24 months) = 24 ÷ 8 = 3 periods.

Effective Compound Interest for 3 periods at 10% = 33.1%.

CI = 33.1% of 5,500 = 0.331 × 5,500 = 1,820.50.

∴ The correct answer is Rs. 1,820.50.

Alternate Method

Given:

Principal (P) = Rs. 5,500

Rate (R) = 15% p.a.

Time (T) = 2 years = 24 months

Compounding Period = Every 8 months

Calculations:

⇒ Effective Rate (r) = (15 ÷ 12) × 8 = 10% per 8-month period.

⇒ Number of periods (n) = 24 months ÷ 8 months = 3.

⇒ Amount (A) = P × (1 + r/100)ⁿ

⇒ A = 5500 × (1 + 10/100)³

⇒ A = 5500 × (1.1)³

⇒ A = 5500 × 1.331

⇒ A = Rs. 7,320.50

⇒ Compound Interest (CI) = A − P

⇒ CI = 7,320.50 − 5,500

⇒ CI = Rs. 1,820.50

∴ The correct answer is Rs. 1,820.50.

Additional Information

Effective Rate for Compounding

When interest is compounded every 'k' months, the new rate r' = (Annual Rate ÷ 12) × k and new periods n' = (Total months ÷ k).

Effective CI Percentage (Successive %)

For 3 periods at r%, the effective percentage is 3r + (3r²/100) + (r³/10000). For 10%, it is 3(10) + 3(100)/100 + 1000/10000 = 33.1%.

Difference between CI and SI

For 2 periods, CI − SI = P(r/100)². For 3 periods, CI − SI = P(r/100)² × (3 + r/100).

Given:

$(2.\bar{4} \times 0.\bar{6} \times 30 \times 0.1\bar{6}) \times [0.2\bar{7} \times (0.8\bar{3} \div 0.1\bar{6})]$

Formula used:

Conversion of recurring decimals to fractions: $0.\bar{a} = \frac{a}{9}$ and $0.a\bar{b} = \frac{ab - a}{90}$

Calculation:

$2.\bar{4} = 2 + \frac{4}{9} = \frac{22}{9}$

⇒ $0.\bar{6} = \frac{6}{9} = \frac{2}{3}$

⇒ $0.1\bar{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}$

⇒ $0.2\bar{7} = \frac{27-2}{90} = \frac{25}{90} = \frac{5}{18}$

⇒ $0.8\bar{3} = \frac{83-8}{90} = \frac{75}{90} = \frac{5}{6}$

⇒ $( \frac{22}{9} \times \frac{2}{3} \times 30 \times \frac{1}{6} ) \times [ \frac{5}{18} \times ( \frac{5}{6} \div \frac{1}{6} ) ]$

⇒ $( \frac{22}{9} \times \frac{2}{3} \times 5 ) \times [ \frac{5}{18} \times 5 ]$

⇒ $\frac{220}{27} \times \frac{25}{18}$

⇒ $\frac{5500}{486} = \frac{2750}{243} \approx 11.3168$

⇒ $\approx 11.3$

∴ The correct answer is option 3

Shortcut Trick

Let Cost Price (CP) of A = 100 units.

A profit = 20% of 100 = 20 units. B CP = 120 units.

B profit = 30% of 120 = 36 units.

Difference in profits = 36 − 20 = 16 units → Given as ₹80.

16 units = ₹80 ⇒ 1 unit = ₹5.

D CP = (120 + 36) × 90% = 156 × 0.9 = 140.4 units → 140.4 × 5 = ₹702.

∴ The correct answer is ₹702.

Alternate Method

Given:

A profit% = 20%

B profit% = 30%

C loss% = 10%

Difference between B's profit and A's profit = ₹80

Formula Used:

Selling Price (SP) = CP × (100 + P%)/100 or CP × (100 − L%)/100

Profit = SP − CP

Calculations:

Let CP for A = 100x

⇒ Profit of A = 20% of 100x = 20x

⇒ SP for A = CP for B = 100x + 20x = 120x

⇒ Profit of B = 30% of 120x = 36x

⇒ According to question: 36x − 20x = 80

⇒ 16x = 80 ⇒ x = 5

⇒ CP for C = SP for B = 120x + 36x = 156x

⇒ CP for D = SP for C = 156x × (100 − 10)/100

⇒ CP for D = 156x × 0.9 = 140.4x

⇒ D bought it for = 140.4 × 5 = ₹702

∴ The correct answer is ₹702.

Additional Information

Successive Profit/Loss Formula

If an object is sold through multiple parties with change percentages a%, b%, and c%, the final SP = Initial CP × (1 ± a/100) × (1 ± b/100) × (1 ± c/100).

Effective Percentage Change

For two successive changes x% and y%, the net change is given by [x + y + (xy/100)]%.

Relationship between SP and CP

SP = CP + Profit or SP = CP − Loss. Profit% is always calculated on the CP unless mentioned otherwise.

Given:

$\left[ \frac{4}{7} \text{ of } 2\frac{4}{5} \times 1\frac{2}{3} - \left( 3\frac{1}{2} - 2\frac{1}{6} \right) \right] \div \left( 3\frac{1}{5} \div 4\frac{1}{2} \text{ of } 5\frac{1}{3} \right)$

Formula used:

BODMAS Rule (Bracket, Of, Division, Multiplication, Addition, Subtraction)

Calculations:

⇒ $\left[ \frac{4}{7} \times \frac{14}{5} \times \frac{5}{3} - \left( \frac{7}{2} - \frac{13}{6} \right) \right] \div \left( \frac{16}{5} \div \left( \frac{9}{2} \times \frac{16}{3} \right) \right)$

⇒ $\left[ \frac{8}{5} \times \frac{5}{3} - \left( \frac{21 - 13}{6} \right) \right] \div \left( \frac{16}{5} \div 24 \right)$

⇒ $\left[ \frac{8}{3} - \frac{8}{6} \right] \div \left( \frac{16}{5} \times \frac{1}{24} \right)$

⇒ $\left[ \frac{8}{3} - \frac{4}{3} \right] \div \frac{2}{15}$

⇒ $\frac{4}{3} \div \frac{2}{15}$

⇒ $\frac{4}{3} \times \frac{15}{2}$ = 2 × 5 = 10

∴ The correct answer is option (1).

Shortcut Trick

Ratio of students in A and B = (90 + 10)/2 : (90 − 10)/2 = 50 : 40 = 5 : 4

Ratio of average scores A : B = 130% : 100% = 13 : 10

Let average score of B be 10k and A be 13k.

Weighted average = (5 × 13k + 4 × 10k) ÷ (5 + 4) = 105k ÷ 9

63 = 105k ÷ 9 ⇒ k = (63 × 9) ÷ 105 = 5.4 → Average of B = 10 × 5.4 = 54

∴ The correct answer is 54.

Alternate Method

Given: Total students = 90, Combined Average = 63, N_A − N_B = 10, Avg_A = 1.3 × Avg_B.

Formula: Combined Average = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂)

⇒ N_A + N_B = 90 and N_A − N_B = 10

⇒ Adding both: 2N_A = 100 ⇒ N_A = 50; N_B = 40

⇒ Let Average of Section B = x, then Average of Section A = 1.3x

⇒ Total Marks = (50 × 1.3x) + (40 × x) = 65x + 40x = 105x

⇒ Combined Average = Total Marks ÷ Total Students

⇒ 63 = 105x ÷ 90

⇒ 105x = 63 × 90 ⇒ x = 5670 ÷ 105 = 54

∴ The correct answer is 54.

Additional Information

Weighted Average Formula

It is the average of two or more groups where each group has a different number of members: x_avg = (n₁x₁ + n₂x₂ + ...) ÷ (n₁ + n₂ + ...).

Alligation Rule

The ratio of the quantities of two components is inversely proportional to the difference between their averages and the combined average: n₁/n₂ = (x₂ − x_avg) ÷ (x_avg − x₁).

Percentage Average Increase

If an average increases by r%, the new average is calculated as New Average = Original Average × (1 + r/100).

Shortcut Trick

Given: CSA = 3696 cm² and h = 3r.

CSA = 2 × π × r × (3r) = 6 × πr² = 3696.

Base Area (πr²) = 3696 ÷ 6 = 616 cm².

πr² = 616 ⇒ (22/7) × r² = 616 ⇒ r² = 196 ⇒ r = 14 cm.

Volume = Base Area × h = 616 × (3 × 14) = 25872 cm³.

Capacity = 25872 ÷ 1000 = 25.872 litres.

∴ The correct answer is 25.872 litres.

Alternate Method

Given:

Curved Surface Area (CSA) = 3696 cm²

Height (h) = 3 × Radius (r)

Formula Used:

Curved Surface Area of Cylinder = 2 × π × r × h

Volume of Cylinder = π × r² × h

1 Litre = 1000 cm³

Calculations:

⇒ 2 × π × r × h = 3696

⇒ 2 × (22/7) × r × 3r = 3696

⇒ (132/7) × r² = 3696

⇒ r² = (3696 × 7) / 132

⇒ r² = 28 × 7 = 196

⇒ r = √196 = 14 cm

⇒ h = 3 × 14 = 42 cm

⇒ Volume = (22/7) × 14 × 14 × 42

⇒ Volume = 22 × 2 × 14 × 42

⇒ Volume = 25872 cm³

⇒ Capacity in litres = 25872 / 1000

⇒ Capacity = 25.872 litres

∴ The correct answer is 25.872 litres.

Additional Information

Total Surface Area of Cylinder

The total surface area includes the curved area and the areas of the two circular bases: TSA = 2πr(h + r).

Volume and Capacity Relation

1 cm³ = 1 mL and 1000 cm³ = 1 Litre. Also, 1 m³ = 1000 Litres = 1 kilolitre.

Volume in terms of CSA

The volume can be expressed as Volume = (CSA × r) ÷ 2. This is useful when radius and CSA are known.

Shortcut Trick

Each exterior angle of a regular polygon is given by 360° ÷ n.

For a regular pentagon (n = 5), each exterior angle = 360° ÷ 5 = 72°.

In the triangle FAE formed by extending the sides, the angles at A and E are the exterior angles.

Sum of angles in ΔFAE = 180°, so ∠EFA = 180° − (72° + 72°).

∠EFA = 180° − 144° = 36°.

∴ The correct answer is 36°.

Alternate Method

Given: A regular pentagon ABCDE where sides BA and DE are produced to meet at point F.

Formula Used: Interior angle of a regular polygon = [(n − 2) × 180°] ÷ n

⇒ Interior angle of pentagon = [(5 − 2) × 180°] ÷ 5

⇒ Interior angle = (3 × 180°) ÷ 5 = 108°

⇒ ∠BAE = 108° and ∠AED = 108°

⇒ Since BAF and DEF are straight lines:

⇒ ∠FAE = 180° − ∠BAE = 180° − 108° = 72°

⇒ ∠FEA = 180° − ∠AED = 180° − 108° = 72°

⇒ In ΔFAE, ∠EFA + ∠FAE + ∠FEA = 180°

⇒ ∠EFA + 72° + 72° = 180°

⇒ ∠EFA + 144° = 180°

⇒ ∠EFA = 180° − 144° = 36°

∴ The correct answer is 36°.

Additional Information

Regular Polygon Angles

For a regular polygon with n sides, the exterior angle is 360° ÷ n and the interior angle is 180° − (360° ÷ n).

Sum of Exterior Angles

The sum of the exterior angles of any convex polygon is always 360°, regardless of the number of sides.

Polygon Interior Sum

The sum of all interior angles of a polygon with n sides is given by (n − 2) × 180°.

Given:

$\frac{15(\sqrt{10} + \sqrt{5})}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$

Formula used:

(a - b)(a + b) = a² - b²

(a + b)² = a² + b² + 2ab

Calculation:

Expression = $\frac{15(\sqrt{10} + \sqrt{5})}{\sqrt{10} + 2\sqrt{5} + 2\sqrt{10} - \sqrt{5} - 4\sqrt{5}}$

⇒ $\frac{15(\sqrt{10} + \sqrt{5})}{3\sqrt{10} - 3\sqrt{5}}$

⇒ $\frac{15(\sqrt{10} + \sqrt{5})}{3(\sqrt{10} - \sqrt{5})}$

⇒ $\frac{5(\sqrt{10} + \sqrt{5})}{\sqrt{10} - \sqrt{5}}$

⇒ $\frac{5(\sqrt{10} + \sqrt{5})}{\sqrt{10} - \sqrt{5}} \times \frac{\sqrt{10} + \sqrt{5}}{\sqrt{10} + \sqrt{5}}$

⇒ $\frac{5(\sqrt{10} + \sqrt{5})^2}{10 - 5}$

⇒ $\frac{5(10 + 5 + 2\sqrt{50})}{5}$

⇒ $15 + 10\sqrt{2}$

⇒ $5(3 + 2\sqrt{2})$

∴ The correct answer is option (1).

Shortcut Trick

Distance covered in first 10 hours = 11.8 × 10 = 118 km.

Remaining distance to be covered = 150 − 118 = 32 km.

Remaining time available = 15 − 10 = 5 hours.

Required speed for remaining part = 32 ÷ 5 = 6.4 km/hr.

∴ The correct answer is 6.4 km/hr.

Alternate Method

Given:

Total Distance = 150 km

Total Time = 15 hours

Initial Speed = 11.8 km/hr for Time = 10 hours

Formula Used:

Distance = Speed × Time

Speed = Distance ÷ Time

⇒ Distance covered in first 10 hours = 11.8 km/hr × 10 hr = 118 km

⇒ Remaining Distance = Total Distance − Distance covered

⇒ Remaining Distance = 150 km − 118 km = 32 km

⇒ Remaining Time = Total Time − Time already spent

⇒ Remaining Time = 15 hr − 10 hr = 5 hours

⇒ Required Speed = Remaining Distance ÷ Remaining Time

⇒ Required Speed = 32 km ÷ 5 hr = 6.4 km/hr

∴ The correct answer is 6.4 km/hr.

Additional Information

Average Speed Formula

Average Speed = Total Distance ÷ Total Time. In this problem, the overall average speed is 150 ÷ 15 = 10 km/hr.

Unit Conversion for Speed

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Constant Distance Relationship

When distance is constant, speed is inversely proportional to time (S₁/S₂ = T₂/T₁).

Shortcut Trick

Surface Area of remaining solid = Curved Surface Area of Cylinder + 2 × Curved Surface Area of Cone

Slant height of cone (l) = √(5² + 12²) = √(25 + 144) = 13 cm

Remaining Area = (2 × π × 5 × 30) + 2 × (π × 5 × 13)

Remaining Area = 300π + 130π = 430π

∴ The correct answer is 430π cm².

Alternate Method

Given:

Height of cylinder (h_c) = 30 cm

Diameter of cylinder = 10 cm → Radius (r) = 5 cm

Radius of conical holes (r) = 5 cm

Height of conical holes (h_k) = 12 cm

Formula Used:

Curved Surface Area (CSA) of Cylinder = 2πrh

Curved Surface Area (CSA) of Cone = πrl

Slant height of cone (l) = √(r² + h²)

Calculations:

⇒ Slant height of cone (l) = √(5² + 12²)

⇒ l = √(25 + 144) = √169 = 13 cm

⇒ CSA of Cylinder = 2 × π × 5 × 30 = 300π cm²

⇒ CSA of one cone = π × 5 × 13 = 65π cm²

⇒ Since holes are drilled out from both bases, original base areas are replaced by conical internal surfaces.

⇒ Total Surface Area = CSA of Cylinder + 2 × (CSA of Cone)

⇒ Total Surface Area = 300π + 2 × 65π

⇒ Total Surface Area = 300π + 130π = 430π cm²

∴ The correct answer is 430π cm².

Additional Information

Cylinder Properties

Volume = πr²h | Total Surface Area = 2πr(h + r) | CSA = 2πrh.

Cone Properties

Volume = (1/3)πr²h | Slant height l = √(r² + h²) | CSA = πrl.

Combined Solids Rule

When a cavity is removed, Volume decreases (V_total − V_removed) but Surface Area often increases as new internal surfaces are exposed.

Given:

9x² + y² = 37

xy = 2

Formula used:

(a + b)² = a² + b² + 2ab

a³ + b³ = (a + b)(a² - ab + b²)

Calculation:

3xy = 3 × 2 = 6

⇒ (3x + y)² = (9x² + y²) + 2(3xy)

⇒ (3x + y)² = 37 + 2(6)

⇒ (3x + y)² = 37 + 12 = 49

⇒ 3x + y = 7

⇒ 27x³ + y³ = (3x + y)((9x² + y²) - 3xy)

⇒ 27x³ + y³ = 7 × (37 - 6)

⇒ 27x³ + y³ = 7 × 31

⇒ 27x³ + y³ = 217

∴ The correct answer is option 4.

Shortcut Trick

Substitute standard trigonometric values: Numerator = 2² × (1/√2)² + 2² = 6.

Denominator = √3 × (√2)² − 2² × 1 = 2√3 − 4.

Divide Numerator by Denominator: 6 / (2√3 − 4) = 3 / (√3 − 2).

Rationalize the denominator: 3(√3 + 2) / (3 − 4) = −3(2 + √3).

∴ The correct answer is -3(2 + √3).

Alternate Method

Given: (cosec² 30° sin² 45° + sec² 60°) ÷ (tan 60° cosec² 45° − sec² 60° tan 45°)

Formula Used: Substitution of standard trigonometric ratios: cosec 30° = 2, sin 45° = 1/√2, sec 60° = 2, tan 60° = √3, cosec 45° = √2, tan 45° = 1.

⇒ Numerator = (2)² × (1/√2)² + (2)²

⇒ Numerator = 4 × (1/2) + 4 = 2 + 4 = 6

⇒ Denominator = (√3) × (√2)² − (2)² × (1)

⇒ Denominator = √3 × 2 − 4 = 2√3 − 4

⇒ Expression = 6 / (2√3 − 4)

⇒ Expression = 3 / (√3 − 2)

⇒ Rationalizing the denominator: [3 × (√3 + 2)] ÷ [(√3 − 2) × (√3 + 2)]

⇒ Expression = [3(√3 + 2)] ÷ (3 − 4) = [3(√3 + 2)] ÷ (−1)

⇒ Expression = −3(2 + √3)

∴ The correct answer is -3(2 + √3).

Additional Information

Standard Trigonometric Values

It is essential to memorize ratios for 0°, 30°, 45°, 60°, and 90°. For example, sin 30° = 1/2, cos 30° = √3/2, and tan 45° = 1.

Reciprocal Ratios

Cosec θ = 1/sin θ, Sec θ = 1/cos θ, and Cot θ = 1/tan θ. Powers of these ratios follow the same reciprocal logic.

Rationalization Rule

To remove a radical from the denominator of a fraction like a/(√b − c), multiply both the numerator and denominator by the conjugate (√b + c).

Shortcut Trick

For similar triangles, the ratio of areas is equal to the square of the ratio of their corresponding sides.

Area(ΔDEF) ÷ Area(ΔABC) = (EF ÷ BC)²

12 ÷ 9 = (EF ÷ 2.1)² ⇒ 4 ÷ 3 = (EF ÷ 2.1)²

Taking square root: EF ÷ 2.1 = 2 ÷ √3 ⇒ EF = 4.2 ÷ √3 = (4.2 × √3) ÷ 3 = 1.4√3 cm.

Since 1.4 = 7/5, EF = 7√3/5 cm.

∴ The correct answer is 7√3/5 cm.

Alternate Method

Given: Area of ΔABC = 9 cm², Area of ΔDEF = 12 cm², Side BC = 2.1 cm.

Formula Used: Ratio of areas of two similar triangles = (Ratio of corresponding sides)².

⇒ Area(ΔDEF) ÷ Area(ΔABC) = (EF ÷ BC)²

⇒ 12 ÷ 9 = (EF ÷ 2.1)²

⇒ 4 ÷ 3 = (EF ÷ 2.1)²

⇒ EF ÷ 2.1 = √(4 ÷ 3) = 2 ÷ √3

⇒ EF = (2.1 × 2) ÷ √3 = 4.2 ÷ √3

⇒ EF = (4.2 ÷ √3) × (√3 ÷ √3) = 4.2√3 ÷ 3

⇒ EF = 1.4 × √3 = (14 ÷ 10) × √3 = 7√3 ÷ 5 cm

∴ The correct answer is 7√3/5 cm.

Additional Information

Ratio of Perimeters

The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides: P₁ ÷ P₂ = s₁ ÷ s₂.

Ratio of Medians and Altitudes

For similar triangles, the ratio of corresponding medians, altitudes, and angle bisectors is also equal to the ratio of their corresponding sides.

Basic Proportionality Theorem (BPT)

If a line is drawn parallel to one side of a triangle to intersect the other two sides, then the other two sides are divided in the same ratio.

Shortcut Trick

Use the basic algebraic expansion: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Substitute the given values: 6² = 50 + 2(ab + bc + ca)

36 = 50 + 2(ab + bc + ca)

2(ab + bc + ca) = 36 − 50 = −14

ab + bc + ca = −14 ÷ 2 = −7

∴ The correct answer is - 7.

Alternate Method

Given: a + b + c = 6, a² + b² + c² = 50, and a³ + b³ + c³ − 3abc = 342

Formula Used: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − (ab + bc + ca))

⇒ 342 = 6 × (50 − (ab + bc + ca))

⇒ 342 ÷ 6 = 50 − (ab + bc + ca)

⇒ 57 = 50 − (ab + bc + ca)

⇒ ab + bc + ca = 50 − 57

⇒ ab + bc + ca = −7

∴ The correct answer is - 7.

Additional Information

Sum of Cubes Identity

The identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). It is also written as ½(a + b + c)[(a − b)² + (b − c)² + (c − a)²].

Zero Sum Property

If a + b + c = 0, then the expression a³ + b³ + c³ − 3abc becomes zero, which implies a³ + b³ + c³ = 3abc.

Square of a Trinomial

The square of the sum of three terms is (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

Shortcut Trick

Divide the equation 3x² − 5x + 1 = 0 by 3x to get x + 1/3x = 5/3.

Square both sides: (x + 1/3x)² = (5/3)².

Apply the identity: x² + 1/9x² + 2 × (1/3) = 25/9.

Subtract 2/3 from both sides: x² + 1/9x² = 25/9 − 6/9 = 19/9.

∴ The correct answer is 19/9.

Alternate Method

Given:

3x² − 5x + 1 = 0

Formula Used:

(a + b)² = a² + b² + 2ab

Calculations:

⇒ 3x² + 1 = 5x

⇒ (3x² + 1) ÷ 3x = 5x ÷ 3x

⇒ (3x² ÷ 3x) + (1 ÷ 3x) = 5/3

⇒ x + 1/3x = 5/3

⇒ Squaring both sides: (x + 1/3x)² = (5/3)²

⇒ x² + (1/3x)² + 2 × x × (1/3x) = 25/9

⇒ x² + 1/9x² + 2/3 = 25/9

⇒ x² + 1/9x² = 25/9 − 2/3

⇒ x² + 1/9x² = (25 − 6) ÷ 9

⇒ x² + 1/9x² = 19/9

∴ The correct answer is 19/9.

Additional Information

Algebraic Expansion

The identity (a + b)² = a² + b² + 2ab is essential for solving problems involving sums of squares of reciprocal terms.

Reciprocal Property

If x + 1/x = k, then x² + 1/x² = k² − 2. This is a common shortcut used in competitive exams.

Generalized Reciprocal Case

If ax + 1/bx = k, then (ax)² + 1/(bx)² = k² − 2 × (a/b). This formula allows direct calculation without full squaring.

Shortcut Trick

Volume of shell = Mass ÷ Density = 6688 ÷ 10.5 = 636.95 ≈ 637 cm³.

Volume formula: (4/3) × π × (R³ − r³) = 637.

External radius R = 6, so (88/21) × (216 − r³) = 637.

216 − r³ = (637 × 21) ÷ 88 = 152.

r³ = 216 − 152 = 64 ⇒ r = 4 cm.

Thickness = R − r = 6 − 4 = 2 cm.

∴ The correct answer is 2 cm.

Alternate Method

Given:

External Radius (R) = 6 cm

Mass of the shell = 6688 g

Density of the metal = 10.5 g/cm³

Formula used: Volume = Mass ÷ Density | Volume of Spherical Shell = (4/3) × π × (R³ − r³)

Calculations:

⇒ Volume of the shell = 6688 ÷ 10.5

⇒ Volume = 6688 ÷ (21/2) = (6688 × 2) ÷ 21 = 13376 ÷ 21 cm³

⇒ (4/3) × (22/7) × (6³ − r³) = 13376 ÷ 21

⇒ (88/21) × (216 − r³) = 13376 ÷ 21

⇒ 88 × (216 − r³) = 13376

⇒ 216 − r³ = 13376 ÷ 88

⇒ 216 − r³ = 152

⇒ r³ = 216 − 152

⇒ r³ = 64

⇒ r = √³64 = 4 cm

⇒ Thickness of the shell = R − r = 6 − 4 = 2 cm

∴ The correct answer is 2 cm.

Additional Information

Volume of a Solid Sphere

The volume of a solid sphere is given by V = (4/3)πr³, where r is the radius.

Surface Area of a Sphere

The total surface area of a sphere is 4πr². For a shell, the external surface area is 4πR².

Density, Mass, and Volume Relation

Density = Mass ÷ Volume. For objects made of a specific material, the mass is proportional to the volume of the material used.

Thickness of Hollow Objects

For any hollow symmetric object, Thickness = External Radius (R) − Internal Radius (r).

Shortcut Trick Difference between numbers = 1280 From division rule: Greater − Smaller = 1280 Given: Greater = 7 × Smaller + 50 Substitute: (7 × Smaller + 50) − Smaller = 1280 ⇒ 6 × Smaller = 1230 Smaller number = 1230 ÷ 6 = 205 Greater number = 205 + 1280 = 1485 ∴ The correct answer is 1485.

Alternate Method Given: Difference (D) = 1280 Quotient (Q) = 7 Remainder (R) = 50 Formula Used: Dividend = (Divisor × Quotient) + Remainder

Let the smaller number be y and the greater number be x. ⇒ x − y = 1280 --- (1) ⇒ x = 7y + 50 --- (2) Substitute the value of x from equation (2) into equation (1): ⇒ (7y + 50) − y = 1280 ⇒ 6y + 50 = 1280 ⇒ 6y = 1280 − 50 ⇒ 6y = 1230 ⇒ y = 1230 ÷ 6 ⇒ y = 205 Substitute y = 205 in equation (1) to find x: ⇒ x − 205 = 1280 ⇒ x = 1280 + 205 ⇒ x = 1485 ∴ The correct answer is 1485.

Additional Information Division Algorithm It states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. Remainder Property The remainder in any division is always non-negative and strictly less than the divisor. If the remainder is 0, the dividend is exactly divisible by the divisor. Linear Equations in Two Variables A system of two linear equations like x − y = k1 and x = my + k2 can be solved efficiently using the substitution or elimination method.

Shortcut Trick

Efficiency ratio A : B : C = 100 : 125 : (125 × 1.4) = 100 : 125 : 175 = 4 : 5 : 7

Total work = Efficiency of A × Days = 4 × 15 = 60 units

Work done by (A + C) in 3 days = (4 + 7) × 3 = 33 units

Remaining work = 60 − 33 = 27 units

Time taken by (A + B) for remaining work = 27 ÷ (4 + 5) = 3 days

∴ The correct answer is 3 days.

Alternate Method

Given:

Time taken by A = 15 days

Efficiency of B = 125% of A

Efficiency of C = 140% of B

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Let Efficiency of A = 4 units/day

⇒ Efficiency of B = 4 × (125/100) = 5 units/day

⇒ Efficiency of C = 5 × (140/100) = 7 units/day

⇒ Total Work = 15 × 4 = 60 units

⇒ Work done by A and C in 3 days = (Efficiency of A + Efficiency of C) × 3

⇒ Work done by A and C = (4 + 7) × 3 = 33 units

⇒ Remaining Work = 60 − 33 = 27 units

⇒ Time taken by A and B = Remaining Work ÷ (Efficiency of A + Efficiency of B)

⇒ Time taken by A and B = 27 ÷ (4 + 5) = 27 ÷ 9 = 3 days

∴ The correct answer is 3 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a work. If the ratio of efficiencies is x : y, the ratio of time taken is y : x.

Work and Wages

Wages are always distributed in the ratio of the work done by each individual, which is equal to the ratio of their efficiencies if they work for the same duration.

Combined Efficiency

When multiple people work together, their net efficiency is the sum of their individual efficiencies. Total Work = (E₁ + E₂ + ... + E_n) × Time.

Shortcut Trick

Volume of Right Prism = Area of Base × Height

Area of Regular Hexagon = (3√3 ÷ 2) × side²

Volume = (3√3 ÷ 2) × 5² × 12√3

Volume = (3√3 ÷ 2) × 25 × 12√3 = 1.5 × 25 × 12 × 3

Volume = 37.5 × 36 = 1350 cm³

∴ The correct answer is 1350 cm³.

Alternate Method

Given:

Side of regular hexagon (a) = 5 cm

Height of prism (h) = 12√3 cm

Formula Used:

Area of regular hexagon = 6 × (√3 ÷ 4) × a²

Volume of prism = Area of base × height

Calculations:

⇒ Area of base = 6 × (√3 ÷ 4) × 5²

⇒ Area of base = 6 × (√3 ÷ 4) × 25

⇒ Area of base = 1.5 × 25 × √3

⇒ Area of base = 37.5√3 cm²

⇒ Volume of prism = 37.5√3 × 12√3

⇒ Volume of prism = 37.5 × 12 × (√3 × √3)

⇒ Volume of prism = 450 × 3

⇒ Volume of prism = 1350 cm³

∴ The correct answer is 1350 cm³.

Additional Information

Volume of Right Prism

The volume is always the product of the base area and the vertical height: V = Area of base × h.

Total Surface Area (TSA) of Prism

TSA = 2 × Base Area + Lateral Surface Area = 2 × Base Area + (Perimeter of base × h).

Regular Hexagon Properties

A regular hexagon consists of 6 equilateral triangles. Area = 6 × (√3 ÷ 4) × side². Perimeter = 6 × side.

Given:

$x = \left( \frac{\sqrt{1875}}{\sqrt{3888}} \div \frac{\sqrt{1200}}{\sqrt{768}} \right) \times \frac{\sqrt{175}}{\sqrt{1792}}$

Formula used:

$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$

Calculations:

x = $\left( \frac{\sqrt{625 \times 3}}{\sqrt{1296 \times 3}} \div \frac{\sqrt{400 \times 3}}{\sqrt{256 \times 3}} \right) \times \frac{\sqrt{25 \times 7}}{\sqrt{256 \times 7}}$

⇒ x = $\left( \frac{25\sqrt{3}}{36\sqrt{3}} \div \frac{20\sqrt{3}}{16\sqrt{3}} \right) \times \frac{5\sqrt{7}}{16\sqrt{7}}$

⇒ x = $\left( \frac{25}{36} \times \frac{16}{20} \right) \times \frac{5}{16}$

⇒ x = $\frac{5}{9} \times \frac{5}{16}$

⇒ x = $\frac{25}{144}$

⇒ $\sqrt{x} = \sqrt{\frac{25}{144}}$

⇒ $\sqrt{x} = \frac{5}{12}$

∴ The correct answer is option (3).

Shortcut Trick

Base Area = πr² = 81π ⇒ Radius (r) = 9 cm.

Height (h) = 12 cm. Using Pythagorean triplet (9, 12, 15), Slant Height (l) = 15 cm.

Curved Surface Area (CSA) = π × r × l

CSA = π × 9 × 15 = 135π cm².

∴ The correct answer is 135π cm².

Alternate Method

Given:

Base Area of Cone = 81π cm²

Height of Cone (h) = 12 cm

Formula Used:

1. Area of circular base = πr²
2. Slant height (l) = √(r² + h²)
3. Curved Surface Area (CSA) = πrl

Calculations:

⇒ πr² = 81π

⇒ r² = 81

⇒ r = √81 = 9 cm

⇒ l = √(r² + h²)

⇒ l = √(9² + 12²)

⇒ l = √(81 + 144) = √225

⇒ l = 15 cm

⇒ Curved Surface Area = π × r × l

⇒ CSA = π × 9 × 15

⇒ CSA = 135π cm²

∴ The correct answer is 135π cm².

Additional Information

Total Surface Area (TSA) of Cone

The TSA includes both the curved surface and the base area: TSA = πrl + πr² = πr(l + r).

Volume of Cone

The volume is one-third the volume of a cylinder with the same radius and height: Volume = (1/3)πr²h.

Slant Height Relationship

The height (h), radius (r), and slant height (l) of a right circular cone form a right-angled triangle, satisfying l² = r² + h².

Shortcut Trick

Identify complementary angle pairs: (38°, 52°), (64°, 26°), and (23°, 67°).

Numerator: 2(sin 38° × sec 52°)² + (cos 64° × sin 26° + sin² 64°) = 2(1) + (cos² 64° + sin² 64°) = 2 + 1 = 3.

Denominator: (tan² 23° − sec² 67°) + (cot² 23° − cosec² 67°) = (tan² 23° − cosec² 23°) + (cot² 23° − sec² 23°)? No.

Correct Denominator: (tan² 23° − cosec² 67°) = (tan² 23° − sec² 23°) = −1 and (cot² 23° − sec² 67°) = (cot² 23° − cosec² 23°) = −1.

Final Result = 3 ÷ (−1 − 1) = 3 ÷ (−2) = −3/2.

∴ The correct answer is −3/2.

Alternate Method

Given expression: (2 sin² 38° sec² 52° + cos 64° sin 26° + sin² 64°) ÷ (tan² 23° + cot² 23° − sec² 67° − cosec² 67°)

Step 1: Simplify the Numerator

⇒ 2 sin² 38° sec² (90° − 38°) + cos 64° sin (90° − 64°) + sin² 64°

⇒ 2 sin² 38° cosec² 38° + cos 64° cos 64° + sin² 64°

⇒ 2 (sin 38° × cosec 38°)² + cos² 64° + sin² 64°

⇒ 2 (1)² + 1 = 3

Step 2: Simplify the Denominator

⇒ tan² 23° + cot² 23° − sec² 67° − cosec² 67°

⇒ tan² 23° + cot² 23° − cosec² (90° − 67°) − sec² (90° − 67°)

⇒ tan² 23° + cot² 23° − cosec² 23° − sec² 23°

⇒ (tan² 23° − sec² 23°) + (cot² 23° − cosec² 23°)

Using identities sec² θ − tan² θ = 1 and cosec² θ − cot² θ = 1:

⇒ (−1) + (−1) = −2

Step 3: Calculate the Final Value

⇒ Value = 3 ÷ (−2) = −3/2

∴ The correct answer is −3/2.

Additional Information

Complementary Angle Trigonometric Ratios

sin (90° − θ) = cos θ | cos (90° − θ) = sin θ | tan (90° − θ) = cot θ | sec (90° − θ) = cosec θ.

Trigonometric Identities

sin² θ + cos² θ = 1 | sec² θ − tan² θ = 1 | cosec² θ − cot² θ = 1.

Reciprocal Ratios

sin θ × cosec θ = 1 | cos θ × sec θ = 1 | tan θ × cot θ = 1.

Given:

The revenue in 2015 and 2017 is 210 crores and 240 crores respectively

The expenditure of the company in 2016, 2018 and 2019 is 175 crores, 165 crores and 170 crores respectively

Concept Used:

Concept of percentage. It is calculated on the base of 100

For Example, x% means x out of 100

Calculation:

The total revenue in 2015 and 2017 is (210 + 240) crores

⇒ 450 crores

The total expenditure of the company in 2016, 2018 and 2019 is (175 + 165 + 170) crores

⇒ 510 crores

The required percentage is {(450/510) × 100}%

⇒ 88.23%

∴ The total revenue in 2015 and 2017 is 88.2% of the total expenditure of the company in 2016, 2018 and 2019 (correct to one decimal place)

Shortcut Trick

Assume lengths $AC = 3$ and $BC = 3$ for simplicity.

Given $AP : PC = 1 : 2$, then $PC = 2$ and $AC = 3$.

Given $BQ : QC = 1 : 2$, then $QC = 2$ and $BC = 3$.

By Pythagoras, $AQ² = 3² + 2² = 13$and$BP² = 3² + 2² = 13$.

Also, $AB² = 3² + 3² = 18$.

Ratio $(AQ² + BP²) ÷ AB² = (13 + 13) ÷ 18 = 26 ÷ 18 = 13/9$.

∴ The correct answer is 13/9.

Alternate Method

Given:

In ΔABC, ∠C = 90°

AP : PC = 1 : 2 ⇒ AC = 3x, PC = 2x

BQ : QC = 1 : 2 ⇒ BC = 3y, QC = 2y

Formula Used:

Pythagoras Theorem: Hypotenuse² = Perpendicular² + Base²

Calculations:

⇒ In right-angled ΔAQC: AQ² = AC² + QC²

⇒ AQ² = (3x)² + (2y)² = 9x² + 4y² --- (i)

⇒ In right-angled ΔBPC: BP² = BC² + PC²

⇒ BP² = (3y)² + (2x)² = 9y² + 4x² --- (ii)

⇒ Adding (i) and (ii): AQ² + BP² = 9x² + 4y² + 9y² + 4x²

⇒ AQ² + BP² = 13x² + 13y² = 13(x² + y²) --- (iii)

⇒ In right-angled ΔABC: AB² = AC² + BC²

⇒ AB² = (3x)² + (3y)² = 9x² + 9y² = 9(x² + y²) --- (iv)

⇒ Dividing (iii) by (iv): (AQ² + BP²) ÷ AB² = [13(x² + y²)] ÷ [9(x² + y²)]

⇒ (AQ² + BP²) ÷ AB² = 13/9

∴ The correct answer is 13/9.

Additional Information

Pythagoras Theorem

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: H² = P² + B².

Median Property in Right Triangle

If AD and BE are medians of ΔABC (right angled at C), then 4(AD² + BE²) = 5AB².

Apollonius Theorem

For any triangle ABC and median AD: AB² + AC² = 2(AD² + BD²).

Shortcut Trick

Let the Cost Price (CP) of the article be 100 units.

Marked Price (MP) = 140% of 100 = 140 units.

Selling Price (SP) = 73.5% of 140 = 0.735 × 140 = 102.9 units.

Profit % = SP − CP = 102.9 − 100 = 2.9%.

∴ The correct answer is 2.9%.

Alternate Method

Given:

Markup percentage = 40% above CP

Selling Price (SP) = 73 1/2% of MP = 73.5% of MP

Formula Used:

MP = CP × (100 + Markup%)/100

SP = MP × (SP percentage of MP)/100

Profit % = [(SP − CP) / CP] × 100

Calculations:

Let CP = 100

⇒ MP = 100 + 40% of 100 = 140

⇒ SP = 73.5% of 140

⇒ SP = (73.5 ÷ 100) × 140

⇒ SP = 0.735 × 140 = 102.9

⇒ Profit = SP − CP = 102.9 − 100 = 2.9

⇒ Profit % = (2.9 ÷ 100) × 100 = 2.9%

∴ The correct answer is 2.9%.

Additional Information

Markup Percentage

It is the percentage by which the cost price is increased to set the marked price. Markup % = [(MP − CP) ÷ CP] × 100.

Discount and SP Relation

Discount is always calculated on the Marked Price. SP = MP − Discount. If SP is given as a percentage of MP, then Discount % = 100% − (SP as % of MP).

Effective Profit Formula

Profit % can also be found using: Profit % = Markup % − Discount % − [(Markup % × Discount %) ÷ 100].

Shortcut Trick

∠BEC and ∠CED form linear pair ⇒ ∠CED = 180° − 128° = 52°

In ▵CDE: ∠CDE = 180° − (52° + 25°) = 103°

∠BAC = ∠BDC = ∠CDE (same arc BC)

∴ The correct answer is 103°.

Alternate Method

Given:

ABCD is a cyclic quadrilateral. The diagonals BD and AC intersect each other at point E. If ∠BEC = 128° and ∠ECD = 25°.

Concept Used:

Angles standing on the same arc in a cyclic quadrilateral are always equal.

Calculation:

∠BEC and ∠CED lie on the same straight line.

∠BEC = 128°

∠CED = 180° - 128°

⇒ ∠CED = 52°

In ΔCDE, ∠CED = 52° and ∠DCE = 25°

∠CDE = 180° - (52° + 25°)

⇒ ∠CDE = 103°

∠BAC and ∠BDC stand on the same arc BC.

We know that angles standing on the same arc in a cyclic quadrilateral are always equal.

∠BAC = 103°

∴ The measure of ∠BAC is 103°.

Additional Information

Angles in the Same Segment

Angles subtended by the same arc at any point on the remaining part of the circle are equal.

Cyclic Quadrilateral Property

The sum of the opposite angles of a cyclic quadrilateral is always 180° (∠A + ∠C = 180° and ∠B + ∠D = 180°).

Exterior Angle Theorem

The exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Shortcut Trick

Total SI rate for 2 years at 12% p.a. = 24%.

Principal (P) = 15500 ÷ (100% + 24%) = 15500 ÷ 1.24 = ₹12,500.

For CI (half-yearly): Rate = 10% / 2 = 5% and Time = 1 1/2 years = 3 periods.

Effective CI rate for 5% for 3 periods = 15.7625%.

Final Amount = 12500 × (100% + 15.7625%) = 12500 × 1.157625 ≈ ₹14,470.

∴ The correct answer is ₹14,470.

Alternate Method

Given:

SI Amount (A_si) = ₹15,500, Time (T_si) = 2 years, Rate (R_si) = 12% p.a.

CI Rate (R_ci) = 10% p.a. (compounded half-yearly), Time (T_ci) = 1 1/2 years.

Formula Used:

Amount in SI = P × (1 + RT/100)

Amount in CI = P × (1 + r/100)ⁿ

Step 1: Find Principal from SI data

⇒ 15500 = P × (1 + (12 × 2)/100)

⇒ 15500 = P × (1 + 0.24)

⇒ 15500 = P × 1.24

⇒ P = 15500 ÷ 1.24 = ₹12,500

Step 2: Calculate CI Amount (compounded half-yearly)

⇒ New rate (r) = 10% ÷ 2 = 5% per half-year

⇒ Time periods (n) = 1.5 × 2 = 3 half-years

⇒ Amount = 12500 × (1 + 5/100)³

⇒ Amount = 12500 × (1.05)³

⇒ Amount = 12500 × 1.157625

⇒ Amount = ₹14,470.3125 ≈ ₹14,470

∴ The correct answer is ₹14,470.

Additional Information

CI Compounding Rules

When compounded half-yearly, the annual rate is halved (R/2) and the time is doubled (2T). For quarterly compounding, the rate is divided by 4 and time is multiplied by 4.

Effective Interest Rate

The effective rate for 3 successive changes of x% is calculated as [3x + 3x²/100 + x³/10000]%. For x=5%, it is 15% + 0.75% + 0.0125% = 15.7625%.

SI vs CI Difference

Difference for 2 years: D = P(R/100)². Difference for 3 years: D = P(R/100)² × (3 + R/100).

Shortcut Trick

Speed ratio S_A : S_B = 120 : 100 = 6 : 5.

Time taken after meeting: T_A = p and T_B = 7.2 hours (since 7 1/5 = 36/5).

Using shortcut: (S_A / S_B)² = T_B / T_A

⇒ (6 / 5)² = 7.2 ÷ p

⇒ 36 / 25 = 7.2 / p ⇒ p = (7.2 × 25) ÷ 36 = 180 ÷ 36 = 5.

∴ The correct answer is 5.

Alternate Method

Given:

Speed of A (S_A) = Speed of B (S_B) × 1.20

Time taken by A after meeting to reach Y (T_A) = p hours

Time taken by B after meeting to reach X (T_B) = 7⅕ hours = 36/5 hours = 7.2 hours

Formula Used:

When two objects start from X and Y simultaneously and move towards each other, after meeting they take T_A and T_B time to reach their destinations, then: S_A / S_B = √(T_B / T_A)

Calculations:

⇒ S_A / S_B = 120 / 100 = 6 / 5

⇒ 6 / 5 = √(7.2 / p)

Squaring both sides:

⇒ (6 / 5)² = 7.2 / p

⇒ 36 / 25 = 7.2 / p

⇒ 36 × p = 7.2 × 25

⇒ 36p = 180

⇒ p = 180 ÷ 36

⇒ p = 5

∴ The correct answer is 5.

Additional Information

Meeting Time Formula

The time 't' elapsed from the start until they meet is given by t = √(T_A × T_B). In this case, t = √(5 × 7.2) = √36 = 6 hours.

Speed-Time Relation

For a fixed distance covered before and after meeting, the ratio of speeds is the square root of the inverse ratio of the times taken to cover the remaining distances.

Total Distance Calculation

Total Distance XY = S_A × (t + T_A) or S_B × (t + T_B). Using the values, Distance = 6k(6 + 5) = 66k or 5k(6 + 7.2) = 5k(13.2) = 66k.

Shortcut Trick

Given sec θ = a/b ⇒ cos θ = b/a.

Express the numerator: 1 − tan²θ = 1 − (sec²θ − 1) = 2 − sec²θ = 2 − (a/b)² = (2b² − a²) / b².

Express the denominator: 2 − sin²θ = 2 − (1 − cos²θ) = 1 + cos²θ = 1 + (b/a)² = (a² + b²) / a².

Required Ratio = [(2b² − a²) / b²] ÷ [(a² + b²) / a²] = a²(2b² − a²) / b²(a² + b²).

∴ The correct answer is a²(2b² − a²) / b²(a² + b²).

Alternate Method

Given: sec θ = a/b

Formula Used: sin θ = Perpendicular / Hypotenuse; cos θ = Base / Hypotenuse; tan θ = Perpendicular / Base

⇒ From the triangle, sec θ = Hypotenuse / Base = a/b.

⇒ Perpendicular = √(a² − b²).

⇒ tan θ = √(a² − b²) / b ⇒ tan²θ = (a² − b²) / b².

⇒ sin θ = √(a² − b²) / a ⇒ sin²θ = (a² − b²) / a².

⇒ Numerator = 1 − [(a² − b²) / b²] = (b² − a² + b²) / b² = (2b² − a²) / b².

⇒ Denominator = 2 − [(a² − b²) / a²] = (2a² − a² + b²) / a² = (a² + b²) / a².

⇒ Expression = [(2b² − a²) / b²] × [a² / (a² + b²)] = a²(2b² − a²) / b²(a² + b²).

∴ The correct answer is a²(2b² − a²) / b²(a² + b²).

Additional Information

Trigonometric Identities

Standard identities: sin²θ + cos²θ = 1 and 1 + tan²θ = sec²θ. These allow for rapid conversion between terms.

Reciprocal Relations

Key reciprocals: secθ = 1/cosθ, cosecθ = 1/sinθ, and cotθ = 1/tanθ.

Pythagorean Triplet Logic

For any right-angled triangle, a² + b² = c², where c is the hypotenuse. In trigonometry, H² = P² + B².

Shortcut Trick

If sin X = cos Y, then X + Y = 90°.

Equating the angles: 3A + (30° − A) = 90° ⇒ 2A = 60° ⇒ A = 30°.

Substitute A = 30° in 2cosecA + tan²2A − cot²A.

Expression = 2cosec 30° + tan² 60° − cot² 30° = 2(2) + (√3)² − (√3)² = 4.

∴ The correct answer is 4.

Alternate Method

Given: sin 3A = cos(30° − A), where A is an acute angle.

Formula Used: cos θ = sin(90° − θ)

⇒ sin 3A = cos(30° − A)

⇒ sin 3A = sin(90° − (30° − A))

⇒ 3A = 90° − 30° + A

⇒ 3A − A = 60°

⇒ 2A = 60°

⇒ A = 30°

Now, calculate the value of 2cosecA + tan²2A − cot²A:

⇒ 2cosec 30° + tan²(2 × 30°) − cot² 30°

⇒ 2cosec 30° + tan² 60° − cot² 30°

⇒ 2 × 2 + (√3)² − (√3)²

⇒ 4 + 3 − 3

⇒ 4

∴ The correct answer is 4.

Additional Information

Complementary Trigonometric Ratios

For any acute angle θ, the ratios of complementary angles are: sin θ = cos(90° − θ), tan θ = cot(90° − θ), and sec θ = cosec(90° − θ).

Standard Trigonometric Values

Key values for this problem: sin 30° = 1/2, cosec 30° = 2, tan 60° = √3, and cot 30° = √3.

Trigonometric Identities

Fundamental square identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Shortcut Trick

For numbers a, b, c, d to be in proportion after adding x: x = (ad − bc) / ((b + c) − (a + d))

Here, a = 9, b = 15, c = 21, d = 31 ⇒ x = (9 × 31 − 15 × 21) / ((15 + 21) − (9 + 31))

x = (279 − 315) / (36 − 40) = −36 / −4 = 9

Mean proportional between (3x − 2) and (5x + 4) = √((3 × 9 − 2) × (5 × 9 + 4)) = √(25 × 49) = 5 × 7 = 35

∴ The correct answer is 35.

Alternate Method

Given: x is added to 9, 15, 21, and 31 such that (9+x), (15+x), (21+x), and (31+x) are in proportion.

Formula: If a, b, c, d are in proportion, then a/b = c/d.

⇒ (9 + x) / (15 + x) = (21 + x) / (31 + x)

⇒ (9 + x)(31 + x) = (21 + x)(15 + x)

⇒ 279 + 31x + 9x + x² = 315 + 15x + 21x + x²

⇒ 279 + 40x = 315 + 36x

⇒ 40x − 36x = 315 − 279

⇒ 4x = 36 ⇒ x = 9

⇒ (3x − 2) = 3(9) − 2 = 25

⇒ (5x + 4) = 5(9) + 4 = 49

⇒ Mean proportional = √(25 × 49) = 5 × 7 = 35

∴ The correct answer is 35.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is the square root of their product: √(a × b).

Third Proportional

If a, b, and c are in continued proportion (a/b = b/c), then c is the third proportional to a and b, given by c = b²/a.

Fourth Proportional

In the proportion a:b :: c:d, d is the fourth proportional to a, b, and c, calculated as d = (b × c) / a.

Properties of Proportion

Product of extremes (a × d) = Product of means (b × c). Also, Componendo and Dividendo states if a/b = c/d, then (a+b)/(a−b) = (c+d)/(c−d).

Shortcut Trick

In ΔPQR, since ∠Q = 90°, P and R are complementary angles.

Using identities: sec P = cosec R, sin P = cos R, and cosec P = sec R.

Substitute these in the expression: [cosec R(cos R + cos R)] / [cosec R(sin R − sec R)]

Simplified form: 2 cos R / (sin R − sec R).

Given cot R = 1/3 (Base/Perpendicular), then sin R = 3/√10, cos R = 1/√10, and sec R = √10.

Result = (2/√10) / (3/√10 − √10) = (2/√10) / (−7/√10) = −2/7.

∴ The correct answer is −2/7.

Alternate Method

Given:

In ΔPQR, ∠Q = 90°

cot R = 1/3

Formula Used:

cot θ = Base / Perpendicular

In a right triangle at Q, P + R = 90° ⇒ sin P = cos R, sec P = cosec R, cosec P = sec R

Calculations:

⇒ cot R = Base / Perpendicular = QR / PQ = 1 / 3

⇒ PR = √(PQ² + QR²) = √(3² + 1²) = √10

⇒ sin R = 3 / √10, cos R = 1 / √10, cosec R = √10 / 3

⇒ Since ∠Q = 90°, P = 90° − R

⇒ sin P = cos R = 1 / √10

⇒ sec P = cosec R = √10 / 3

⇒ cosec P = sec R = √10 / 1 = √10

⇒ Numerator = sec P(cos R + sin P) = (√10 / 3) × (1/√10 + 1/√10) = (√10 / 3) × (2/√10) = 2 / 3

⇒ Denominator = cosec R(sin R − cosec P) = (√10 / 3) × (3/√10 − √10) = (√10 / 3) × ((3 − 10)/√10) = −7 / 3

⇒ Required Value = (2 / 3) / (−7 / 3) = −2 / 7

∴ The correct answer is −2/7.

Additional Information

Complementary Angle Identities

In a right-angled triangle, if ∠A + ∠B = 90°, then sin A = cos B, tan A = cot B, and sec A = cosec B.

Pythagorean Identities

Fundamental identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Relations

Trigonometric reciprocals are sin θ = 1/cosec θ, cos θ = 1/sec θ, and tan θ = 1/cot θ.